POJ 1113 wall (凸包）

解题思路:

 凸包的周长加上原的周长，最后四舍五入。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define LL long long
using namespace std;
const int MAXN = 1000 + 10;
struct Point
{
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
};
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }
bool operator < (const Point& a, const Point& b)
{
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double eps = 1e-10;
int dcmp(double x)
{
if(fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b)
{
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(Vector A, Vector B) { return A.x * B.x + A.y * B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A,B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; }
int ConvexHull(Point* p, int n, Point* ch)
{
sort(p, p+n);
int m = 0;
for(int i=0;i<n;i++)
{
while(m > 1 && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;
ch[m++] = p[i];
}
int k = m;
for(int i=n-2;i>=0;i--)
{
while(m > k && Cross(ch[m-1] - ch[m-2], p[i] - ch[m-2]) <= 0) m--;
ch[m++] = p[i];
}
if(n > 1) m--;
return m;
}
//Point p[MAXN], ch[MAXN];
double dis(Point A, Point B)
{
double x = A.x - B.x;
double y = A.y - B.y;
return (double) sqrt(x *x + y * y);
}
const double PI = acos(-1);
int main()
{
int N , L;
while(scanf("%d%d", &N, &L)!=EOF)
{
Point p[MAXN], ch[MAXN];
for(int i=0;i<N;i++)
scanf("%lf%lf", &p[i].x, &p[i].y);
int m = ConvexHull(p, N, ch);
double ans = 0.0;
for(int i=0;i<m;i++)
{
if(i != m -1) ans += dis(ch[i], ch[i+1]);
else ans += dis(ch[i], ch[0]);
}
ans += (double) 2 * PI * L;
printf("%d\n", (int) (ans + 0.5));
}
return 0;
}